HDU3572_Task Schedule(网络流最大流)

解题报告

题意：

工厂有m台机器，需要做n个任务。对于一个任务i，你需要花费一个机器Pi天，而且，开始做这个任务的时间要>=Si，完成这个任务的时间<=Ei。对于一个任务，只能由一个机器来完成，一个机器同一时间只能做一个任务。但是，一个任务可以分成几段不连续的时间来完成。问，能否做完全部任务。

思路：

网络流在于建模，这题建模方式是：

把每一天和每个任务看做点。由源点到每一任务，建容量为pi的边（表示任务需要多少天完成）。每个任务到每一天，若是可以在这天做任务，建一条容量为1的边，最后，把每天到汇点再建一条边容量m（表示每台机器最多工作m个任务）。

#include <map>#include <queue>#include <vector>#include <cstdio>#include <cstring>#include <iostream>#define inf 99999999using namespace std;int n,m,l[2010],head[2010],cnt,M;struct node{    int v,w,next;} edge[555000];void add(int u,int v,int w){    edge[M].v=v;    edge[M].w=w;    edge[M].next=head[u];    head[u]=M++;    edge[M].v=u;    edge[M].w=0;    edge[M].next=head[v];    head[v]=M++;}int bfs(){    memset(l,-1,sizeof(l));    l[0]=0;    int i,u,v;    queue<int >Q;    Q.push(0);    while(!Q.empty())    {        u=Q.front();        Q.pop();        for(i=head[u]; i!=-1; i=edge[i].next)        {            v=edge[i].v;            if(l[v]==-1&&edge[i].w>0)            {                l[v]=l[u]+1;                Q.push(v);            }        }    }    if(l[cnt]>0)return 1;    return 0;}int dfs(int u,int f){    int a,i;    if(u==cnt)return f;    for(i=head[u]; i!=-1; i=edge[i].next)    {        int v=edge[i].v;        if(l[v]==l[u]+1&&edge[i].w>0&&(a=dfs(v,min(f,edge[i].w))))        {            edge[i].w-=a;            edge[i^1].w+=a;            return a;        }    }    l[u]=-1;//没加优化会T    return 0;}int main(){    int t,i,j,s,p,e,k=1;    scanf("%d",&t);    while(t--)    {        M=0;        memset(head,-1,sizeof(head));        scanf("%d%d",&n,&m);        int sum=0,maxx=0;        for(i=1; i<=n; i++)        {            scanf("%d%d%d",&p,&s,&e);            add(0,i,p);            sum+=p;            if(maxx<e)                maxx=e;            for(j=s; j<=e; j++)                add(i,j+n,1);        }        cnt=n+maxx+1;        for(i=1; i<=maxx; i++)        {            add(n+i,cnt,m);        }        int ans=0,a;        while(bfs())            while(a=dfs(0,inf))                ans+=a;        printf("Case %d: ",k++);        if(ans==sum)            printf("Yes\n");        else printf("No\n");        printf("\n");    }    return 0;}

Task Schedule

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3311    Accepted Submission(s): 1154

Problem Description

Our geometry princess XMM has stoped her study in computational geometry to concentrate on her newly opened factory. Her factory has introduced M new machines in order to process the coming N tasks. For the i-th task, the factory has to start processing it at or after day Si, process it for Pi days, and finish the task before or at day Ei. A machine can only work on one task at a time, and each task can be processed by at most one machine at a time. However, a task can be interrupted and processed on different machines on different days. 
Now she wonders whether he has a feasible schedule to finish all the tasks in time. She turns to you for help.

 

Input

On the first line comes an integer T(T<=20), indicating the number of test cases.

You are given two integer N(N<=500) and M(M<=200) on the first line of each test case. Then on each of next N lines are three integers Pi, Si and Ei (1<=Pi, Si, Ei<=500), which have the meaning described in the description. It is guaranteed that in a feasible schedule every task that can be finished will be done before or at its end day.

 

Output

For each test case, print “Case x: ” first, where x is the case number. If there exists a feasible schedule to finish all the tasks, print “Yes”, otherwise print “No”.

Print a blank line after each test case.

 

Sample Input

24 31 3 5 1 1 42 3 73 5 92 22 1 31 2 2

 

Sample Output

Case 1: Yes   Case 2: Yes

 

Author

allenlowesy

 

Source

2010 ACM-ICPC Multi-University Training Contest（13）——Host by UESTC